# Confusion about mean curvature and variation of surface area

For a two-dimensional surface in $\mathbb{R}^3$, I thought that the total mean curvature was equal to the first variation of area:

$$\frac{d}{dt}SA(t)\Big\vert_{t\to 0} = \int H dA,$$

where $SA(t)$ is the surface area of the surface after flowing it along its normal vector field for time $t$.

But when I try this formula for a cylinder of unit height and radius $r$, I get that $SA(t) = 2\pi (r+t)$, $H=\frac{1}{2r}$, and

$$2\pi \stackrel{?}{=} 2\pi r \frac{1}{2r} = \pi.$$

Where have I gone wrong? Am I missing a factor of two in the first variation of area formula?

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If $n$ is an unit normal field to the surface $S$ then, for any $\phi\in C_c^\infty(S),$ you should get $\left.\dfrac{d}{dt}\right|_{t=0}\textrm{Surface}(\{x+t\phi(x) n(x):x\in S\})=\mathbf{2}\int_S\phi(x)H(x)d\sigma(x).$ – Giuseppe Jun 12 '12 at 7:25
Ah! You're right, it looks like the paper I read used the annoying alternate definition of "mean" curvature as $H=k_1+k_2$. Thanks! – user7530 Jun 12 '12 at 7:44
You could write that up as an answer and accept it so the question doesn't remain unanswered. – joriki Jun 12 '12 at 9:49

$$\int 2 H dA.$$
Sometimes in the literature mean curvature is defined as $H = k_1 + k_2$ instead of $(k_1+k_2)/2$ and this discrepancy was the source of my error.